The half life period of a reaction is halved as the initial concentration of the reactants is doubled. The order of reaction is:

Options

A. 0.5
B. 2
C. 1
D. zero

Answer

B

Explanation

c) `t_(1//2) propto 1/([A_(0)]^(n-1)),(t_(1//2))_(1)/(t_(1//2))_(2) = [A_(0)]_(2)^(n-1)/([A_(0)]_(1)^(n-1))={([A_(0)]_(2))/([A_(0)]_(1))}^(n-1)`
`t/(t_(1//2))=(2a)/(a)^(n-1)` or `2=(2)^(n-1)` or n-1 or n=2